Vector Product

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Consider, now, the *cross product* or *vector product*:

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Thus, the -component of transforms correctly. It can easily be shown that the other components transform correctly as well, and that all components also transform correctly under rotation about and . Thus, is a proper vector. Incidentally, is the

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The cross product transforms like a vector, which
means that it must have a well-defined direction and magnitude. We can show
that
is *perpendicular* to both and .
Consider
. If this is zero then the cross product
must be perpendicular to . Now,

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Therefore, is perpendicular to . Likewise, it can be demonstrated that is perpendicular to . The vectors , , and form a

Let us now evaluate the magnitude of
. We have

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Thus,

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Consider the parallelogram defined by the vectors and --see Figure A.103.
The *scalar area* of the parallelogram is
. By convention, the *vector area* has the magnitude of the
scalar area, and is normal to the plane of the parallelogram, in the sense obtained from a right-hand circulation rule by rotating on to
(through an acute angle): *i.e.*, if the fingers of the right-hand circulate in the direction of
rotation then the thumb of the right-hand indicates the direction of the vector area. So, the vector area is coming out of the
page in Figure A.103.
It follows that

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Suppose that a force is applied at position --see Figure A.104.
The torque about the origin is the product of the magnitude of the force and
the length of the lever arm . Thus, the magnitude of the torque is
. The direction of the torque is conventionally defined as the direction of
the axis through about which the force tries to rotate objects, in the sense
determined by a right-hand circulation rule. Hence, the torque is out of the page in Figure A.104.
It follows that the vector torque is
given by

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The angular momentum, , of a particle of linear momentum and position vector is simply defined as the moment of its
momentum about the origin: *i.e.*,

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